Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;
public class Main {
    public static double dp[][] = new double[3005][3005];
    public static double arr[];
    public static double solve(int i , int x){
        if(x == 0) return 1;
        if(i == 0) return 0;

        // System.out.println( i + " " + x);

        if(dp[i][x] > -0.9) return dp[i][x];

        return dp[i][x] = (arr[i-1] * solve(i - 1 , x - 1)) + ((1 - arr[i-1]) * solve(i-1 , x));
    }
    public static void main(String args[]) {
        Scanner s = new Scanner(System.in);
        int n = s.nextInt();

        arr = new double[n];

        for(int i = 0 ; i < n ; i++){
            arr[i] = s.nextDouble();
        }
        for(int i = 0 ; i < dp.length ; i++){
            for(int j = 0 ; j < dp[0].length ; j++){
                dp[i][j] = -1;
            }
        }

        System.out.println(solve(n , (n+1)/2));
    }
}